Question

Trigonometric State the three basic trigonometric ratios for the following

Answer 1

Given

$\beta=-\frac{7\pi}{2}$

To find the three basic trigonometric ratios.

Explanation;

It is given that,

$\beta=-\frac{7\pi}{2}$

That implies,

$\begin{gathered} \sin\beta=sin(-\frac{7\pi}{2}) \\ =-\sin(\frac{7\pi}{2}) \\ =-sin(\frac{6\pi}{2}+\frac{\pi}{2}) \\ =-\sin(3\pi+\frac{\pi}{2}) \\ =-(-sin(\frac{\pi}{2})) \\ =sin\frac{\pi}{2} \\ =1 \end{gathered}$

Also,

$\begin{gathered} \cos\beta=cos(-\frac{7\pi}{2}) \\ =\cos\frac{7\pi}{2} \\ =-cos\frac{\pi}{2} \\ =0 \end{gathered}$

And,

$\begin{gathered} \tan\beta=tan(-\frac{7\pi}{2}) \\ =-tan(\frac{7\pi}{2}) \\ =-tan(\frac{\pi}{2}) \\ =-\infty \end{gathered}$

Hence, the trigonometric ratios are,

$sin\beta=1,cos\beta=0,tan\beta=-\infty$